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Question Number 132082 by liberty last updated on 11/Feb/21

$$\mathrm{Solve}\int \frac{\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^3}?$$

Answered by EDWIN88 last updated on 11/Feb/21

$$\mathrm{Ostrogradski}\mathrm{method}$$$$\mathrm{Consider}\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{ax}}^3+\mathrm{bx}}{{({\mathrm{x}}^2+1)}^2}\right)=\frac{-{\mathrm{ax}}^4+(3\mathrm{a}-3\mathrm{b}){\mathrm{x}}^2+\mathrm{b}}{{({\mathrm{x}}^2+1)}^3}$$$$\int \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{{\mathrm{ax}}^3+\mathrm{bx}}{{({\mathrm{x}}^2+1)}^2}\right]=\int \frac{-{\mathrm{ax}}^4+(3\mathrm{a}-3\mathrm{b}){\mathrm{x}}^2+\mathrm{b}}{{({\mathrm{x}}^2+1)}^3}\mathrm{dx}+\int \frac{\mathrm{c}}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$\mathrm{comparing}\mathrm{coefficients}$$$$1=-{\mathrm{ax}}^4+(3\mathrm{a}-3\mathrm{b}){\mathrm{x}}^2+\mathrm{c}{({\mathrm{x}}^2+1)}^2+\mathrm{b}$$$$1=(\mathrm{c}-\mathrm{a}){\mathrm{x}}^4+(3\mathrm{a}-3\mathrm{b}+2\mathrm{c}){\mathrm{x}}^2+\mathrm{b}+\mathrm{c}$$$$\mathrm{we}\mathrm{get}\mathrm{a}=\mathrm{c},\mathrm{b}+\mathrm{c}=1,3\mathrm{a}-3\mathrm{b}+2\mathrm{c}=0$$$$\mathrm{then}\mathrm{a}=\mathrm{c}=\frac{3}{8};\mathrm{b}=\frac{5}{8}$$$$\mathrm{therefore}\mathrm{I}=\frac{1}{8}\left[\frac{{3\mathrm{x}}^3+5\mathrm{x}}{{({\mathrm{x}}^2+1)}^2}\right]+\frac{3}{8}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+1}$$$$\mathrm{I}=\frac{1}{8}\left[\frac{{3\mathrm{x}}^3+5\mathrm{x}}{{({\mathrm{x}}^2+1)}^2}\right]+\frac{3}{8}\arctan \left(\mathrm{x}\right)+\mathrm{c}$$$$\mathrm{please}\mathrm{check}$$

Answered by liberty last updated on 11/Feb/21

$$\mathrm{by}\mathrm{Ostrogradski}\mathrm{method}$$$$\mathrm{let}\int \frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^3}=\frac{{\mathrm{ax}}^3+\mathrm{bx}}{{({\mathrm{x}}^2+1)}^2}+\int \frac{\mathrm{c}}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$\mathrm{differentiating}\mathrm{both}\mathrm{sides}$$$$\frac{1}{{({\mathrm{x}}^2+1)}^3}=\frac{({3\mathrm{ax}}^2+\mathrm{b}){({\mathrm{x}}^2+1)}^2-4\mathrm{x}({\mathrm{x}}^2+1)({\mathrm{ax}}^3+\mathrm{bx})}{{({\mathrm{x}}^2+1)}^4}+\frac{\mathrm{c}}{{\mathrm{x}}^2+1}$$$$\frac{1}{{({\mathrm{x}}^2+1)}^3}=\frac{({3\mathrm{ax}}^2+\mathrm{b})({\mathrm{x}}^2+1)-({4\mathrm{ax}}^4+{4\mathrm{bx}}^2)}{{({\mathrm{x}}^2+1)}^3}+\frac{\mathrm{c}{({\mathrm{x}}^2+1)}^2}{{({\mathrm{x}}^2+1)}^3}$$$$1={3\mathrm{ax}}^4+{3\mathrm{ax}}^2+{\mathrm{bx}}^2+\mathrm{b}-{4\mathrm{ax}}^4-{4\mathrm{bx}}^2+{\mathrm{cx}}^4+{2\mathrm{cx}}^2+\mathrm{c}$$$$1=(-\mathrm{a}+\mathrm{c}){\mathrm{x}}^4+(3\mathrm{a}-3\mathrm{b}+2\mathrm{c}){\mathrm{x}}^2+\mathrm{b}+\mathrm{c}$$$$\mathrm{we}\mathrm{get}\mathrm{c}=\mathrm{a};\mathrm{b}+\mathrm{c}=1;3\mathrm{a}-3\mathrm{b}+2\mathrm{c}=0$$$$\Rightarrow 5\mathrm{c}-3(1-\mathrm{c})=0;8\mathrm{c}=3$$$$\mathrm{c}=\mathrm{a}=\frac{3}{8}\mathrm{and}\mathrm{b}=\frac{5}{8}$$$$\mathrm{then}\mathrm{the}\mathrm{integral}\mathrm{become}$$$$\int \frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^3}=\frac{{3\mathrm{x}}^3+5\mathrm{x}}{8{({\mathrm{x}}^2+1)}^2}+\int \frac{3}{8}\frac{\mathrm{dx}}{({\mathrm{x}}^2+1)}$$$$=\frac{{3\mathrm{x}}^3+5\mathrm{x}}{8{({\mathrm{x}}^2+1)}^2}+\frac{3}{8}\arctan \mathrm{x}+\mathrm{c}$$$$$$

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